# Need help with calculus homework

asked Aug 2, 2014 in CALCULUS

f (2) = - 3, g(2) = 3, f ' (2) = - 2, and g ' (2) = 5.

a). h(x) = f(x) * g(x)

To find, h ' (2), differentiate the above with respect to x.

h ' (x) = [f(x) * g(x)] '

Apply product rule : [f(x) * g(x)] ' = f (x) g' (x) + g (x)f ' (x)

h ' (x) = f (x) g' (x) + g (x)f ' (x)

h ' (2) = f (2) g' (2) + g (2)f ' (2)

Substitute corresponding values.

h ' (2) = [(- 3)*5] + [3*(-2)]

= - 15 - 6

= - 21.

Therefore, h' (2) = - 21.

b). h (x) = g (x) / (1 + f (x)).

Differentiate the above equation with respect to x.

h '(x) = [ g (x) / (1 + f (x)) ] '

Apply quotient rule : [f(x) / g(x)] ' = [ g(x) f '(x) - f(x)g '(x) ] /  [g(x)]2 .

h '(x) = [ [1 + f(x)]g '(x) - g(x)[1 + f(x)] ' ] / [1 + f(x)]2

h '(x) = [ [1 + f(x)]g '(x) - g(x)f '(x) ] / [1 + f(x)]2

h '(2) = [ [1 + f(2)]g '(2) - g(2)f '(2) ] / [1 + f(2)]2

Substitute corresponding values.

h '(2) = [ [1 + (- 3)]5 - (3)(- 2) ] / [1 + (-3)]2

= [ [1 - 3]5 + 6 ] / [1 - 3]2

= [ [- 2]5 + 6 ] / [- 2]2

= [ - 10 + 6 ] / 4

= - 4/4

= -1.

Therefore, h' (2) = - 1.

3)

a)

Given function y = (1+2x)^16

Take derivative bothe sides with respect to x.

dy/dx =d/dx( (1+2x)^16)

Using formula:d/dx(x^n)=nx^n-1

dy/dx =16 (1+2x)^15 d/dx (1+2x)

=16(1+2x)^15 (2)

=32(1+2x)^15

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b)

Given function y = sin√x

Take derivative bothe sides with respect to x.

dy/dx =d/dx(sin√x)

Using formula:d/dx(x^n)=nx^n-1 and d/dx(sinx)=cosx

dy/dx =cos √x d/dx (√x)

=cos √x (1/2√x)

=(cos√x)/2√x

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c)

Given function y =1/ √(x^2+16)

Take derivative bothe sides with respect to x.

dy/dx =d/dx(1/ √(x^2+16))

Using formula:d/dx(x^n)=nx^n-1

dy/dx = (-1/2)(x^2+16)^3/2*d/dx (x^2+16)

=(-1/2)(x^2+16)^3/2*2x

=-((x^2+16)^3/2)/2

3)

a)

x²y+y³ =3x

Use implicit differention.

d/dx(x²y+y³) =d/dx(3x)

d/dx(x²y)+d/dx(y³) =d/dx(3x)

x²d/dx(y)+yd/dx(x²)+d/dx(y³) =d/dx(3x)

x²dy/dx+y(2x)+3y²dy/dx=3

(x²+3y²)dy/dx=3-2xy

dy/dx=(3-2xy)/(x²+3y²)

--------------------------

b)

sin(x+y) =x+y²

Use implicit differention.

d/dx(sin(x+y)) =d/dx(x+y²)

cos(x+y)d/dx(x+y)=1+2ydy/dx

cos(x+y)(1+dy/dx) =1+2ydy/dx

cos(x+y)+cos(x+y)dy/dx =1+2ydy/dx

cos(x+y)dy/dx-2ydy/dx=1-cos(x+y)

(cos(x+y)-2y)dy/dx=1-cos(x+y)

dy/dx=(1-cos(x+y))/(cos (x+y)-2y)